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Question

Find the vector equation of the plane which is at a distance of 7 units from the origin and normal to the vector 3ˆi+5ˆj6ˆk.

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Solution

The normal vector is n=3ˆi+5ˆj6ˆk
Therefore,
ˆn=n|n|=3ˆi+5ˆj6ˆk32+52+62=3ˆi+5ˆj6ˆk70

Equation of the plane with postion vector n is given by, r.ˆn=d

ˆr.(3ˆi+5ˆj6ˆk70)=7

This is the vector equation of the required plane.

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